Problem: Recently you have bought a snow walking robot and brought it home. Suppose your home is a cell (0, 0) on an infinite grid.

You also have the sequence of instructions of this robot. It is written as the string s consisting of characters 'L', 'R', 'U' and 'D'. If the robot is in the cell (x, y) right now, he can move to one of the adjacent cells (depending on the current instruction).

  * If the current instruction is 'L', then the robot can move to the left to (x - 1, y); 
  * if the current instruction is 'R', then the robot can move to the right to (x + 1, y); 
  * if the current instruction is 'U', then the robot can move to the top to (x, y + 1); 
  * if the current instruction is 'D', then the robot can move to the bottom to (x, y - 1). 



You've noticed the warning on the last page of the manual: if the robot visits some cell (except (0, 0)) twice then it breaks.

So the sequence of instructions is valid if the robot starts in the cell (0, 0), performs the given instructions, visits no cell other than (0, 0) two or more times and ends the path in the cell (0, 0). Also cell (0, 0) should be visited at most two times: at the beginning and at the end (if the path is empty then it is visited only once). For example, the following sequences of instructions are considered valid: "UD", "RL", "UUURULLDDDDLDDRRUU", and the following are considered invalid: "U" (the endpoint is not (0, 0)) and "UUDD" (the cell (0, 1) is visited twice).

The initial sequence of instructions, however, might be not valid. You don't want your robot to break so you decided to reprogram it in the following way: you will remove some (possibly, all or none) instructions from the initial sequence of instructions, then rearrange the remaining instructions as you wish and turn on your robot to move. 

Your task is to remove as few instructions from the initial sequence as possible and rearrange the remaining ones so that the sequence is valid. Report the valid sequence of the maximum length you can obtain.

Note that you can choose any order of remaining instructions (you don't need to minimize the number of swaps or any other similar metric).

You have to answer q independent test cases.

Input

The first line of the input contains one integer q (1 ≤ q ≤ 2 ⋅ 10^4) — the number of test cases.

The next q lines contain test cases. The i-th test case is given as the string s consisting of at least 1 and no more than 10^5 characters 'L', 'R', 'U' and 'D' — the initial sequence of instructions.

It is guaranteed that the sum of |s| (where |s| is the length of s) does not exceed 10^5 over all test cases (∑ |s| ≤ 10^5).

Output

For each test case print the answer on it. In the first line print the maximum number of remaining instructions. In the second line print the valid sequence of remaining instructions t the robot has to perform. The moves are performed from left to right in the order of the printed sequence. If there are several answers, you can print any. If the answer is 0, you are allowed to print an empty line (but you can don't print it).

Example

Input


6
LRU
DURLDRUDRULRDURDDL
LRUDDLRUDRUL
LLLLRRRR
URDUR
LLL


Output


2
LR
14
RUURDDDDLLLUUR
12
ULDDDRRRUULL
2
LR
2
UD
0

Note

There are only two possible answers in the first test case: "LR" and "RL".

The picture corresponding to the second test case:

<image> Note that the direction of traverse does not matter 

Another correct answer to the third test case: "URDDLLLUURDR".

Solution (in JAVA):

Solution:
import java.io.BufferedReader;
import java.io.FileReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashMap;
import java.util.HashSet;
import java.util.LinkedList;
import java.util.Queue;
import java.util.StringTokenizer;
import java.util.TreeSet;

public class Main {
	static  ArrayList<Pair>adj[];
	static int memo2[];
	static char c1[][];
	static HashMap<Long,Integer> hs=new HashMap<>();
	static PrintWriter out=new PrintWriter(System.out);
	public static void main(String[] args) throws Exception {
	Scanner sc=new Scanner(System.in);
	int n=sc.nextInt();
	while(n-->0){
	String s=sc.nextLine();
	int up=0;
	int down=0;
	int left=0;
	int right=0;
	char c[]=s.toCharArray();
	for (int i = 0; i < c.length; i++) {
		if(c[i]=='U') {
			up++;
		}
		else if(c[i]=='D') {
			down++;
		}
		else if(c[i]=='R') {
			right++;
		}
		else if(c[i]=='L') {
			left++;
		}
	}
	int dir1=Math.min(up, down);
	int dir2=Math.min(left, right);
	StringBuilder sb=new StringBuilder();
	for (int i = 0; i <dir1; i++) {
		sb.append('U');
		
	}
	for (int i = 0; i <dir2; i++) {
		sb.append('R');
		
	}
	for (int i = 0; i <dir1; i++) {
		sb.append('D');
		
	}
	for (int i = 0; i <dir2; i++) {
		sb.append('L');
		
	}
	if(dir1==0&&dir2!=0) {
		System.out.println(2);
		System.out.println("LR");
	}
	else if(dir2==0&&dir1!=0) {
		System.out.println(2);
		System.out.println("UD");
	}
	else {
	System.out.println(sb.length());
	System.out.println(sb);
	}
	}
	}
	static  class Triple implements Comparable<Triple>{
		int cost;
		int totalcost;
		int idx;
		public Triple(int a,int b,int c) {
			cost=b;
			totalcost=c;
			idx=a;
		}
		@Override
		public int compareTo(Triple o) {
		if(this.totalcost!=o.totalcost)
			return this.totalcost-o.totalcost;
		else
			return this.cost-o.cost;
			
		}
		
	}
	private static int dp(int cur, int size) {
	if(size==k) {
		return 1;
	}
	if(memo[cur][size]!=-1) {
		return memo[cur][size];
	}
	long count=0;
	for(int i=1 ; i*cur<=n ; i++) {
			count+=dp(i*cur,size+1);
		
	}
	return memo[cur][size]=(int) count;
	}
	static TreeSet<Long> primeFactors(long N)		// O(sqrt(N) / ln sqrt(N))
	{
		TreeSet<Long> factors = new TreeSet<Long>();		//take abs(N) in case of -ve integers
		int idx = 0, p = primes.get(idx);

		while(p * p <= N)
		{
			while(N % p == 0) { factors.add((long) p); N /= p; }
			if(primes.size()>idx+1)
			p = primes.get(++idx);
			else
				break;
		}

		if(N != 1)						// last prime factor may be > sqrt(N)
			factors.add(N);				// for integers whose largest prime factor has a power of 1
		return factors;
	}
	
	static boolean visited[];
	static int k;
	static int bfs(int s)
	{
		Queue<Integer> q = new LinkedList<Integer>();
		q.add(s);
		int count=0;
		int maxcost=0;
		int dist[]=new int[n];
		dist[s]=0;
		while(!q.isEmpty())
		{
			
			int u = q.remove();
			if(dist[u]==k) {
				break;
			}
			for(Pair v: adj[u])
			{	
				maxcost=Math.max(maxcost, v.cost);
			
			
			
			if(!visited[v.v])	{
				
				visited[v.v]=true;
				q.add(v.v);
				dist[v.v]=dist[u]+1;
				maxcost=Math.max(maxcost, v.cost);
			}
			}	
			
		}
		return maxcost;
	}
	public static boolean FindAllElements(int n, int k) {
	 int sum = k;  
     int[] A = new int[k];  
      Arrays.fill(A, 0, k, 1);  
        
      for (int i = k - 1; i >= 0; --i) {  
      
           while (sum + A[i] <= n) {  
      
              sum += A[i];  
              A[i] *= 2;  
          }  
      }  
      if(sum==n) {
      	return true;
      }
      else
    	return false; 
	}
	
	static int[][] memo;
	static long b[];

	static class Pair implements Comparable<Pair>{
		int v;
		int cost;
		public Pair(int x , int y) {
		v=x;
		cost=y;
		}
		@Override
		public int compareTo(Pair o) {
		return o.cost-this.cost;
		}
		
	}
	static long bs(long x) {
		long high=(long) 1e7;
		long low=0;
		long idxend=0;
		long idxstart=0;
		if(hs.containsKey(x)) {
			return hs.get(x);
		}
		while(high>=low) {
			long mid=(high+low)>>1;
			long range=((mid)*(mid-1));
			range/=2;
			range++;
			if(range>x) {
				high=mid-1;
				
			}
			else {
				low=mid+1;
			idxstart=mid;
			}
		}
	return idxstart;	
	}
	static boolean vis2[][];
	
	static class SegmentTree {		// 1-based DS, OOP
		
		int N; 			//the number of elements in the array as a power of 2 (i.e. after padding)
		int[] array, sTree, lazy;
		
		SegmentTree(int[] in)		
		{
			array = in; N = in.length - 1;
			sTree = new int[N<<1];		//no. of nodes = 2*N - 1, we add one to cross out index zero
			lazy = new int[N<<1];
			build(1,1,N);
		}
		
		void build(int node, int b, int e)	// O(n)
		{
			if(b == e)					
				sTree[node] = array[b];
			else						
			{
				int mid = b + e >> 1;
				build(node<<1,b,mid);
				build(node<<1|1,mid+1,e);
				sTree[node] = sTree[node<<1]+sTree[node<<1|1];
			}
		}
		
		
		void update_point(int index, int val)			// O(log n)
		{
			index += N - 1;				
			sTree[index] += val;			
			while(index>1)				
			{
				index >>= 1;
				sTree[index] = sTree[index<<1] + sTree[index<<1|1];		
			}
		}
		
		
		void update_range(int i, int j, int val)		// O(log n) 
		{
			update_range(1,1,N,i,j,val);
		}
		
		void update_range(int node, int b, int e, int i, int j, int val)
		{
			if(i > e || j < b)		
				return;
			if(b >= i && e <= j)		
			{
				sTree[node] += (e-b+1)*val;			
				lazy[node] += val;				
			}							
			else		
			{
				int mid = b + e >> 1;
				propagate(node, b, mid, e);
				update_range(node<<1,b,mid,i,j,val);
				update_range(node<<1|1,mid+1,e,i,j,val);
				sTree[node] = sTree[node<<1] + sTree[node<<1|1];		
			}
			
		}
		void propagate(int node, int b, int mid, int e)		
		{
			lazy[node<<1] += lazy[node];
			lazy[node<<1|1] += lazy[node];
			sTree[node<<1] += (mid-b+1)*lazy[node];		
			sTree[node<<1|1] += (e-mid)*lazy[node];		
			lazy[node] = 0;
		}
		
		int query(int i, int j)
		{
			return query(1,1,N,i,j);
		}
		
		int query(int node, int b, int e, int i, int j)	// O(log n)
		{
			if(i>e || j <b)
				return 0;		
			if(b>= i && e <= j)
				return sTree[node];
			int mid = b + e >> 1;
			propagate(node, b, mid, e);
			int q1 = query(node<<1,b,mid,i,j);
			int q2 = query(node<<1|1,mid+1,e,i,j);
			return q1 + q2;	
					
		}
	
	}
	static long mod=(long) (1e9+7);
	static boolean f2=false;
	static long[][] matMul(long[][] a2, long[][] b, int p, int q, int r)	//C(p x r) = A(p x q) x (q x r) -- O(p x q x r)
	{
		long[][] C = new long[p][r];
		for(int i = 0; i < p; ++i) {
			for(int j = 0; j < r; ++j) {
				for(int k = 0; k < q; ++k) {
					C[i][j] = (C[i][j]+(a2[i][k]%mod * b[k][j]%mod))%mod;
		 C[i][j]%=mod;
				}
		
			}
		}
		return C;
	}
		static long query[];
	static long last1=0;
	static TreeSet<Long> x1;
	static HashMap<Long,Long> map;
	static long max=0;
	static int first=0;
	
	
	
		
		
		
		

		
	
	static ArrayList<Pair> a1[];
	static int memo1[];
	
	static boolean vis[];
	static TreeSet<Integer> set=new TreeSet<Integer>();
	
	static long modPow(long ways, long count, long mod)	// O(log e)
	{
		ways %= mod;
		long res = 1;
		while(count > 0)
		{
			if((count & 1) == 1)
				res = (res * ways) % mod;
			ways = (ways * ways) % mod;
			count >>= 1;
		}
		return res%mod;
	}
	
	
	
	static  int  dx[] = {0,0,0,3,3,3,6,6,6};
	static int dy[] = {0,3,6,0,3,6,0,3,6};
	static long[] a;
	static int n;
	static int s;
	static boolean f[];

	static long gcd(long ans,long b) {
		if(b==0) {
			return ans;
		}
		return gcd(b,ans%b);
	}
	static int[] isComposite;
	static int[] valid;
	
	static ArrayList<Integer> primes;
	static ArrayList<Integer> l;
	static void sieve(int N)	// O(N log log N) 
	{
		isComposite = new int[N+1];					
		isComposite[0] = isComposite[1] = 1;			// 0 indicates a prime number
		primes = new ArrayList<Integer>();
		l=new ArrayList<>();
		valid=new int[N+1];
		for (int i = 2; i <= N; ++i) 					//can loop till i*i <= N if primes array is not needed O(N log log sqrt(N)) 
			if (isComposite[i] == 0) 					//can loop in 2 and odd integers for slightly better performance
			{
				primes.add(i);
				l.add(i);
				valid[i]=1;
				
				for (int j = i*2; j <=N; j +=i) {	// j = i * 2 will not affect performance too much, may alter in modified sieve
						isComposite[j] = 1;
						
						
						
				      
					}
			}
		
		for(int i=0 ; i<primes.size() ; i++) {
			for(int j:primes) {
				if(primes.get(i)*j>N) {
					break;
				}
				valid[primes.get(i)*j]=1;
			}
		}
	}
	static int c;
	static class UnionFind {                                              
		int[] p, rank, setSize;
		int numSets;

		public UnionFind(int N) 
		{
			p = new int[numSets = N];
			rank = new int[N];
			setSize = new int[N];
			for (int i = 0; i < N; i++) {  p[i] = i; setSize[i] = 1; }
		}

		public int findSet(int i) { return p[i] == i ? i : (p[i] = findSet(p[i])); }

		public boolean isSameSet(int i, int j) { return findSet(i) == findSet(j); }

		public void unionSet(int i, int j) 
		{ 
			if (isSameSet(i, j)) 
				return;
			numSets--; 
			int x = findSet(i), y = findSet(j);
			if(rank[x] > rank[y]) { p[y] = x; setSize[x] += setSize[y]; }
			else
			{	p[x] = y; setSize[y] += setSize[x];
				if(rank[x] == rank[y]) rank[y]++; 
			} 
		}

		public int numDisjointSets() { return numSets; }

		public int sizeOfSet(int i) { return setSize[findSet(i)]; }
	}
	public static int[] schuffle(int[] a2) {
		for (int i = 0; i < a2.length; i++) {
			int x=(int)(Math.random()*a2.length);
		int temp=a2[x];
		a2[x]=a2[i];
		a2[i]=temp;
		}
		return a2;
	}
		static int V;
	static int INF=(int) 1E9;
		
		static class Edge2 
		{
			int node;
		long cost;
			long next;
			Edge2(int a, int c,Long long1) { node = a;	cost = long1; next=c;}
			
		}
static class Scanner {
		StringTokenizer st;
		BufferedReader br;
 
		public Scanner(InputStream system) {
			br = new BufferedReader(new InputStreamReader(system));
		}
 
		public Scanner(String file) throws Exception {
			br = new BufferedReader(new FileReader(file));
		}
 
		public String next() throws IOException {
			while (st == null || !st.hasMoreTokens())
				st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}
 
		public String nextLine() throws IOException {
			return br.readLine();
		}
 
		public int nextInt() throws IOException {
			return Integer.parseInt(next());
		}
 
		public double nextDouble() throws IOException {
			return Double.parseDouble(next());
		}
 
		public char nextChar() throws IOException {
			return next().charAt(0);
		}
 
		public Long nextLong() throws IOException {
			return Long.parseLong(next());
		}
 
		public boolean ready() throws IOException {
			return br.ready();
		}
 
		public void waitForInput() throws InterruptedException {
			Thread.sleep(3000);
		}
	}	
}